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2922. Market Analysis III

Description

Table: Users

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| seller_id      | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+
seller_id is column of unique values for this table.
This table contains seller id, join date, and favorite brand of sellers.

Table: Items

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| item_id       | int     |
| item_brand    | varchar |
+---------------+---------+
item_id is the column of unique values for this table.
This table contains item id and item brand.

Table: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| item_id       | int     |
| seller_id     | int     |
+---------------+---------+
order_id is the column of unique values for this table.
item_id is a foreign key to the Items table.
seller_id is a foreign key to the Users table.
This table contains order id, order date, item id and seller id.

Write a solution to find the top seller who has sold the highest number of unique items with a different brand than their favorite brand. If there are multiple sellers with the same highest count, return all of them.

Return the result table ordered by seller_id in ascending order.

The result format is in the following example.

 

Example 1:

Input: 
Users table:
+-----------+------------+----------------+
| seller_id | join_date  | favorite_brand |
+-----------+------------+----------------+
| 1         | 2019-01-01 | Lenovo         |
| 2         | 2019-02-09 | Samsung        |
| 3         | 2019-01-19 | LG             |
+-----------+------------+----------------+
Orders table:
+----------+------------+---------+-----------+
| order_id | order_date | item_id | seller_id |
+----------+------------+---------+-----------+
| 1        | 2019-08-01 | 4       | 2         |
| 2        | 2019-08-02 | 2       | 3         |
| 3        | 2019-08-03 | 3       | 3         |
| 4        | 2019-08-04 | 1       | 2         |
| 5        | 2019-08-04 | 4       | 2         |
+----------+------------+---------+-----------+
Items table:
+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1       | Samsung    |
| 2       | Lenovo     |
| 3       | LG         |
| 4       | HP         |
+---------+------------+
Output: 
+-----------+-----------+
| seller_id | num_items |
+-----------+-----------+
| 2         | 1         |
| 3         | 1         |
+-----------+-----------+
Explanation: 
- The user with seller_id 2 has sold three items, but only two of them are not marked as a favorite. We will include a unique count of 1 because both of these items are identical.
- The user with seller_id 3 has sold two items, but only one of them is not marked as a favorite. We will include just that non-favorite item in our count.
Since seller_ids 2 and 3 have the same count of one item each, they both will be displayed in the output.

Solutions

Solution 1: Equijoin + Grouping + Subquery

We can use equijoin to connect the Orders table and the Users table according to seller_id, then connect Items according to item_id, and filter out the records where item_brand is not equal to favorite_brand. Then, group by seller_id and count the number of item_id corresponding to each seller_id. Finally, use a subquery to find the seller_id with the most item_id.

• SQL

• # Write your MySQL query statement below
  WITH
      T AS (
          SELECT seller_id, COUNT(DISTINCT item_id) AS num_items
          FROM
              Orders
              JOIN Users USING (seller_id)
              JOIN Items USING (item_id)
          WHERE item_brand != favorite_brand
          GROUP BY 1
      )
  SELECT seller_id, num_items
  FROM T
  WHERE num_items = (SELECT MAX(num_items) FROM T)
  ORDER BY 1;
  
  

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